Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 11

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MATH101 C April 2024

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Other MATH101 C Exams

• April 2024 •

Question 11
How many intervals are needed to guarantee that the Simpson's Rule approximation of the integral ${\begin{aligned}I=\int _{0}^{5}f(x)\mathrm {d} x\end{aligned}}$ lies within $0.0004$ of the exact value? Assume that ${\begin{aligned}\|f^{(4)}(x)\|\leq 9\qquad {\rm {for\;all}}\;x\;{\rm {obeying}}\;0\leq x\leq 5,\end{aligned}}$ and note that $0.0004={\frac {1}{20\cdot 5^{3}}}$ .

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Hint
Recall the error estimate formula for Simpson's Rule ${\begin{aligned}\|I-S_{n}\|\leq {\frac {L}{180}}{\frac {(b-a)^{5}}{n^{4}}},\end{aligned}}$ where $n$ must be an even number. Using this formula, we can set up an inequality to ensure the error is within the specified tolerance and solve for $n$ .

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Solution
Let $S_{n}$ denote the Simpson approximation with $n$ subintervals. The inequality in the statement supports the choice $L=9$ in the well-known error estimate formula. This leads to ${\begin{aligned}\|I-S_{n}\|\leq {\frac {L}{180}}{\frac {(b-a)^{5}}{n^{4}}}={\frac {9}{180}}{\frac {(5)^{5}}{n^{4}}}.\end{aligned}}$ To use this, we chain the desired tolerance onto the right side of the inequality, obtaining ${\begin{aligned}{\frac {1}{20}}{\frac {(5)^{5}}{n^{4}}}\leq {\frac {1}{20\cdot 5^{3}}}.\end{aligned}}$ Multiplying both sides by $20(5)^{3}n^{4}$ gives ${\begin{aligned}{\frac {5^{5}}{20}}\cdot 20\cdot 5^{3}\leq n^{4},\quad \mathrm {i.e.,} \quad 5^{8}\leq n^{4}.\end{aligned}}$ This is equivalent to $n\geq 5^{2}=25.$ Since the number of subintervals used in Simpson's Rule must be even, then any even $n\geq 26$ will achieve the specified tolerance.